A 25.00 mL volume of 1.637 M sulfuric acid is titrated with potassium hydroxide. At the equivalence point, 64.78mL of base were added. What is the molarity of the base?
A 25.00 mL volume of 1.637 M sulfuric acid is titrated with potassium hydroxide. At the equivalence point, 64.78mL of base were added. What is the molarity of the base?
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
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![**Titration Problem: Determining the Molarity of a Base**
A 25.00 mL volume of 1.637 M sulfuric acid is titrated with potassium hydroxide. At the equivalence point, 64.78 mL of base were added. What is the molarity of the base?
**[Answer Box]**
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This question involves a titration process where sulfuric acid (H₂SO₄) is neutralized by potassium hydroxide (KOH). To solve this, consider the reaction:
\[ H_2SO_4 + 2KOH \rightarrow K_2SO_4 + 2H_2O \]
From this balanced equation, it is evident that one mole of sulfuric acid reacts with two moles of potassium hydroxide.
Given:
- Volume of H₂SO₄ = 25.00 mL = 0.02500 L
- Molarity of H₂SO₄ = 1.637 M
- Volume of KOH at equivalence point = 64.78 mL = 0.06478 L
To find the molarity of KOH, use the formula:
\[ M_1V_1 = M_2V_2 \]
\[ (1.637 \, \text{M})(0.02500 \, \text{L}) = M_{\text{KOH}}(0.06478 \, \text{L}) \]
\[ M_{\text{KOH}} = \frac{(1.637 \, \text{M})(0.02500 \, \text{L})}{(0.06478 \, \text{L})} = ? \]
Solve for the molarity of the potassium hydroxide solution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F19b0aed3-745e-4879-8577-ea35399669d0%2F0af5fb0f-6a7d-4327-9c7f-9e815f9ec413%2Fp3cl1p7_processed.png&w=3840&q=75)
Transcribed Image Text:**Titration Problem: Determining the Molarity of a Base**
A 25.00 mL volume of 1.637 M sulfuric acid is titrated with potassium hydroxide. At the equivalence point, 64.78 mL of base were added. What is the molarity of the base?
**[Answer Box]**
[Include a space here for the answer box to be placed when implemented on the website]
This question involves a titration process where sulfuric acid (H₂SO₄) is neutralized by potassium hydroxide (KOH). To solve this, consider the reaction:
\[ H_2SO_4 + 2KOH \rightarrow K_2SO_4 + 2H_2O \]
From this balanced equation, it is evident that one mole of sulfuric acid reacts with two moles of potassium hydroxide.
Given:
- Volume of H₂SO₄ = 25.00 mL = 0.02500 L
- Molarity of H₂SO₄ = 1.637 M
- Volume of KOH at equivalence point = 64.78 mL = 0.06478 L
To find the molarity of KOH, use the formula:
\[ M_1V_1 = M_2V_2 \]
\[ (1.637 \, \text{M})(0.02500 \, \text{L}) = M_{\text{KOH}}(0.06478 \, \text{L}) \]
\[ M_{\text{KOH}} = \frac{(1.637 \, \text{M})(0.02500 \, \text{L})}{(0.06478 \, \text{L})} = ? \]
Solve for the molarity of the potassium hydroxide solution.
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